STRUCTURE OF Ni-56 AND Ni-58
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. STRUCTURE OF Ni-56 WITH S = 0 After a detailed analysis of the structure of Ni-56 having 28 protons and 28 neutrons I discovered that it consists of the structure of Mg-24 (from p1 to n12) in which we add 8 additional alpha particles. In the following diagram one sees the structure of Mg-24 with the two additional alpha particles of nucleons as p25, n25, p26, and n26 as well as of the p27, n27, p28, and n28. Note that the 12 nucleons ( from p13 to n18) of the three alpha particles existing in front of Mg-24 are not shown, because they are in front of Mg-24. Similarly the three rest alpha particles existing behind the Mg-24 have the 12 nucleons from p19 to n24. For example as you can see in the first real horizontal plane, it is not the simple square of the structure of Mg-24 but it contains also the deuteron n13p13 existing in front of it and the deuteron n19p19 existing behind it. Although the structure has a high symmetry here it is unstable because of the large number of protons, which exert strong repulsions able to overcome the pn bonds. Under this condition it decays to Co-56 with S=0 . Especially the p28 of Ni-56 decays to n29 of Co-56. (See my STRUCTURE OF Co-59 AND Co-56 ). STRUCTURE OF Ni-56 OF HIGH SYMMETRY WITH S = 0 (The 24 nucleons from p13 to n24 are not shown, because they exist in front of the Mg-24 and behind it) ' ' ' p12..........n12' ' n11..........p11 Sixth horizontal plane' ' n10..........p10' ' p9...........n9 Fifth horizontal plane' ' n26.........p8............n8............p28' ' p26..........n7...........p7...........n28 Fourth horizontal plane' ' p25.........n6...........p6............n27' ' n25……….p5...........n5…………p27 Third horizontal plane' ' p4...........n4' ' n3………..p3 Second horizontal plane' ' n2……….p2' ' p1...........n1 First horizontal plane' ' First real horizontal plane with 8 nucleons' ' p19……..n19' ' n2……….p2' ' p1………n1' ' n13…….p13' STRUCTURE OF Ni-58 WITH S = 0 Using the structure of Ni-56 with S=0 it is possible to get the structure of Ni-58 with S =0 by adding the two neutrons as n29(-1/2) '''and '''n30 (+1/2). They contribute to the increase of the pn bonds per proton able to overcome the pp repulsions of long range . In the following diagram you see that the n29 and n30 '''fill the blank positions formed by the protons of the two alpha particles and the core of the structure of Mg-24. For example the n30(+1/2) fills the blank position formed by the p9 and p26. Thus in the presence of p30 we observe at p9 not only the bonds as (p9-n9), (p9-n11), (p9-n10), and (p9-n7) but also the additional bond as (p9-n30) which contributes to the increase of the binding energies able to overcome the pp repulsions of long range. '''STRUCTURE OF Ni-58 OF HIGH SYMMETRY WITH S = 0 (The 24 nucleons from p13 to n24 are not shown, because they exist in front of the Mg-24 and behind it) ' ' ' p12..........n12' ' n11..........p11 Sixth horizontal plane' ' n10..........p10' ' n30……… p9...........n9 Fifth horizontal plane' ' n26.........p8............n8............p28' ' p26..........n7...........p7...........n28 Fourth horizontal plane' ' p25.........n6...........p6............n27' ' n25……….p5...........n5…………p27 Third horizontal plane' ' p4...........n4' ' n3………..p3…………n29 Second horizontal plane' ' n2………p2' ' p1..........n1 First horizontal plane' ' ' ' ' Category:Fundamental physics concepts